What is the answer to solvefor y,x!=2^y+2^z ?

The answer to solvefor y,x!=2^y+2^z is y=(ln(x!-2^z))/(ln(2))

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solvefor y,x!=2^y+2^z

solve for

y, x! = 2^{y} + 2^{z}

y = \frac{ln ( x ! - 2 ^{z} )}{ln ( 2 )}

Solution steps

x! = 2^{y} + 2^{z}

Switch sides

2^{y} + 2^{z} = x!

Move

2^{z}

to the right side

2^{y} = x! - 2^{z}

Apply exponent rules

y ln (2) = ln (x! - 2^{z})

Solve

y ln (2) = ln (x! - 2^{z}) : y = \frac{ln ( x ! - 2 ^{z} )}{ln ( 2 )}

y = \frac{ln ( x ! - 2 ^{z} )}{ln ( 2 )}

3 6^{x - 1} = \frac{1}{216}

3^{5 x + 1} = 9

(\frac{1}{6})^{x} = 9

4 \cdot 1 0^{2 x} + 1 = 21

x, e^{l n (x)} = e^{y + c}

What is the answer to solvefor y,x!=2^y+2^z ?
The answer to solvefor y,x!=2^y+2^z is y=(ln(x!-2^z))/(ln(2))